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Question

A 0.35 kg0.35\ \text{kg} mass at the end of a spring oscillates 2.52.5 times per second with an amplitude of 0.15 m0.15\ \text{m}. Determine

(a) the velocity when it passes the equilibrium point,

(b). the velocity when it is 0.10 m0.10\ \text{m} from equilibrium,

(c). the total energy of the system, and

(d). the equation x(t)x(t) describing the motion of the mass, assuming that at t=0t=0, x(t)x(t) was a maximum.

Solution

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Answered 7 months ago
Answered 7 months ago
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Let mm be the mass of the block, ff the frequency of oscillation, and AA the amplitude.

m=0.35 kgf=2.5 HzA=0.15 m\begin{align*} m &= 0.35\ \text{kg} \\ f &= 2.5\ \text{Hz} \\ A &= 0.15\ \text{m} \end{align*}

The law of conservation of energy will be applied extensively in this problem.

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